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[SOLVED] Can anyone tell me the use of these resistors in this circuit??

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saluvs

saluvs

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Hi
I got here an automatic battery charger circuit.I dont understand the use of those 10 ohm 10 W parallel combination of resistors used in it..
Can anyone please help me with this??
Thankyou.
 

It's a strange design, the 555 is only being used as a comparator.

The resistors are there to limit the maximum current to the battery. Consider what would happen if a completely discharged (0V) battery was connected, the transformer and bridge rectifier would be driving in to 0V so far too much current would try to flow and damage would occur. The resistors are there to keep the current within safe limits, they drop the charge voltage, converting it into heat until the battery voltage has started to rise.

Brian.
 
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    saluvs

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    peterkos2

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Should the relay coil be connected to the other side of the 10 ohm resistors? Especially if the battery is greatly discharged? Closing the relay contact could re-open it and cause cycling. Is this a problem or a feature?
 

No, I think it's correct. To be honest I would have to look up the internal diagram of the 555 to be absolutely sure but the way I think it works is the 555 output goes high when the power is applied, this closes the relay and charges the battery through the resistors. When the threshold voltage is reached, the 555's internal flip-flop makes the output go low and the relay opens. It's running in monostable mode with no timing capacitor so I think it operates once and has to be powered down to reset it.

I wouldn't be too happy about the battery still being connected to the 555 when the relay opens but it should survive. Perhaps a diode, anode end to the battery + side and cathode to the positive supply would be a good idea. It would prevent the supply to the 555 being removed while other pins still had 12V on them if the power is turned off.

Brian.
 

betwixt said:
No, I think it's correct. To be honest I would have to look up the internal diagram of the 555 to be absolutely sure but the way I think it works is the 555 output goes high when the power is applied, this closes the relay and charges the battery through the resistors. When the threshold voltage is reached, the 555's internal flip-flop makes the output go low and the relay opens. It's running in monostable mode with no timing capacitor so I think it operates once and has to be powered down to reset it.

I wouldn't be too happy about the battery still being connected to the 555 when the relay opens but it should survive. Perhaps a diode, anode end to the battery + side and cathode to the positive supply would be a good idea. It would prevent the supply to the 555 being removed while other pins still had 12V on them if the power is turned off.

Brian.
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My concern is when the relay closes and power is applied to a drained battery, the battery will drag down the coil voltage and the relay will open up again. Then, the voltage goes back up and the relay closes, then opens, etc. This will be less likely if the relay coil was connected on the other side of the 10 ohm resistors.
 
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There is a risk of that happening. If in doubt connect it to the other side of the resistors. It should not make any difference to the operation of the circuit although the relay coil voltage will be a bit higher.

Brian.
 
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